Testing the Difference
Between Two Means:
Dependent Samples
Sec 9.3
Bluman, Chapter 9
1
This the last day the class meets before
spring break starts.
Please make sure to be present for the test
or make appropriate arrangements to take
the test before leaving for spring break
Bluman, Chapter 9
2
9.3 Testing the Difference Between
Two Means: Dependent Samples
When the values are dependent, do a t test on the
differences. Denote the differences with the symbol D, the
mean of the population differences with μD, and the
sample standard deviation of the differences with sD.
D  D
t
sD n
with d.f. = n -1 and where
D

D
n
and
sD 
n D 2    D 2 
Bluman, Chapter 9
n  n  1
3
Chapter 9
Testing the Difference Between
Two Means, Two Proportions,
and Two Variances
Section 9-3
Example 9-6
Page #493
Bluman, Chapter 9
4
Example 9-6: Vitamin for Strength
A physical education director claims by taking a special
vitamin, a weight lifter can increase his strength. Eight
athletes are selected and given a test of strength, using
the standard bench press. After 2 weeks of regular
training, supplemented with the vitamin, they are tested
again. Test the effectiveness of the vitamin regimen at
α = 0.05. Each value in the data represents the maximum
number of pounds the athlete can bench-press. Assume
that the variable is approximately normally distributed.
Bluman, Chapter 9
5
Example 9-6: Vitamin for Strength
Step 1: State the hypotheses and identify the claim.
H0: μD = 0 and H1: μD < 0 (claim)
Step 2: Find the critical value.
The degrees of freedom are n – 1 = 8 – 1 = 7.
The critical value for a left-tailed test with
α = 0.05 is t = -1.895.
Bluman, Chapter 9
6
Example 9-6: Vitamin for Strength
Step 3: Compute the test value.
Before (X1) After (X2)
210
230
182
205
262
253
219
216
219
236
179
204
270
250
222
216
D 19

D

 2.375
n
D2
-9
-6
3
1
-8
3
-3
0
81
36
9
1
64
9
9
0
Σ D = -19
Σ D2 = 209
8
n D    D 
2
sD 
D = X1 – X2
n  n  1
2

8  209   19 
87
Bluman, Chapter 9
2
 4.84
7
Example 9-6: Vitamin for Strength
Step 3: Compute the test value.
D  2.375, sD  4.84
t
2.375  0
D  D

 1.388
4.84 8
sD n
Step 4: Make the decision. Do not reject the null.
Step 5: Summarize the results.
There is not enough evidence to support the
claim that the vitamin increases the strength of
weight lifters.
Bluman, Chapter 9
8
Chapter 9
Testing the Difference Between
Two Means, Two Proportions,
and Two Variances
Section 9-3
Example 9-7
Page #495
Bluman, Chapter 9
9
Example 9-7: Cholesterol Levels
A dietitian wishes to see if a person’s cholesterol level will
change if the diet is supplemented by a certain mineral.
Six subjects were pretested, and then they took the
mineral supplement for a 6-week period. The results are
shown in the table. (Cholesterol level is measured in
milligrams per deciliter.) Can it be concluded that the
cholesterol level has been changed at α = 0.10? Assume
the variable is approximately normally distributed.
Bluman, Chapter 9
10
Example 9-7: Cholesterol Levels
Step 1: State the hypotheses and identify the claim.
H0: μD = 0 and H1: μD  0 (claim)
Step 2: Find the critical value.
The degrees of freedom are 5. At α = 0.10, the
critical values are ±2.015.
Bluman, Chapter 9
11
Example 9-7: Cholesterol Levels
Step 3: Compute the test value.
Before (X1) After (X2)
210
235
208
190
172
244
20
65
-2
2
-1
400
4225
4
4
1
256
210
188
173
228
16
Σ D = 100 Σ D2 = 4890
6
n D    D 
2
sD 
D2
190
170
D 100

D

 16.7
n
D = X1 – X2
n  n  1
2

6  4890  100 
65
Bluman, Chapter 9
2
 25.4
12
Example 9-7: Cholesterol Levels
Step 3: Compute the test value.
D  16.7, sD  25.4
t
16.7  0
D  D

 1.610
sD n 25.4 6
Step 4: Make the decision. Do not reject the null.
Step 5: Summarize the results.
There is not enough evidence to support the
claim that the mineral changes a person’s
cholesterol level.
Bluman, Chapter 9
13
Confidence Interval for the Mean
Difference
Formula for the t confidence interval for the mean
difference
D  t
2
sD
  D  D  t
n
2
sD
n
d.f.  n  1
Bluman, Chapter 9
14
Chapter 9
Testing the Difference Between
Two Means, Two Proportions,
and Two Variances
Section 9-3
Example 9-8
Page #498
Bluman, Chapter 9
15
Example 9-8: Confidence Intervals
Find the 90% confidence interval for the difference
between the means for the data in Example 9–7.
D  t
2
16.7  2.015 
sD
  D  D  t
n
2
sD
n
25.4
25.4
  D  16.7  2.015 
6
6
16.7  20.89  D  16.7  20.89
4.19  D  37.59
Since 0 is contained in the interval, the decision is to
not reject the null hypothesis H0: μD = 0.
Bluman, Chapter 9
16
On your own

Study the examples in
section 9.3


Sec 9.3 page 500
#1-9 odds
Bluman, Chapter 9
17